Let \(\overline{\mathrm{A}}=2 \hat{i}+\hat{k}, \overline{\mathrm{~B}}=\hat{i}+\hat{j}+\hat{k}\) and \(\overline{\mathrm{C}}=4 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+7 \hat{\mathrm{k}}\). If a vector \(\bar{R}\) satisfies \(\bar{R} \times \bar{B}=\bar{C} \times \bar{B}\) and \(\overline{\mathrm{R}} \cdot \overline{\mathrm{A}}=0\), then \(\overline{\mathrm{R}}\) is given by

\(\begin{aligned} & \overline{\mathrm{R}} \times \overline{\mathrm{B}}=\overline{\mathrm{C}} \times \overline{\mathrm{B}} \\ & \Rightarrow \overline{\mathrm{A}} \times(\overline{\mathrm{R}} \times \overline{\mathrm{B}})=\overline{\mathrm{A}} \times(\overline{\mathrm{C}} \times \overline{\mathrm{B}}) \\ & \Rightarrow(\overline{\mathrm{A}} \cdot \overline{\mathrm{B}}) \overline{\mathrm{R}}-(\overline{\mathrm{A}} \cdot \overline{\mathrm{R}}) \overline{\mathrm{B}}=(\overline{\mathrm{A}} \cdot \overline{\mathrm{B}}) \overline{\mathrm{C}}-(\overline{\mathrm{A}} \cdot \overline{\mathrm{C}}) \overline{\mathrm{B}} \\ & \Rightarrow(\overline{\mathrm{A}} \cdot \overline{\mathrm{B}}) \overline{\mathrm{R}}-0=(\overline{\mathrm{A}} \cdot \overline{\mathrm{B}}) \overline{\mathrm{C}}-(\overline{\mathrm{A}} \cdot \overline{\mathrm{C}}) \overline{\mathrm{B}}\end{aligned}\)
\(\begin{aligned}
& \Rightarrow \overline{\mathrm{R}}=\overline{\mathrm{C}}-\left(\frac{\overline{\mathrm{A}} \cdot \overline{\mathrm{C}}}{\overline{\mathrm{~A}} \cdot \overline{\mathrm{~B}}}\right) \overline{\mathrm{B}} \\
& \overline{\mathrm{~A}} \cdot \overline{\mathrm{C}}=(2 \hat{\mathrm{i}}+\hat{\mathrm{k}}) \cdot(4 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+7 \hat{\mathrm{k}}) \\
& =2(4)+0(-3)+1(7) \\
& =15 \\
& \overline{\mathrm{~A}} \cdot \overline{\mathrm{~B}}=(2 \hat{\mathrm{i}}+\hat{\mathrm{k}}) \cdot(\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}) \\
& =2(1)+0(1)+1(1) \\
& =3 \\
& \therefore \quad \overline{\mathrm{R}}=4 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+7 \hat{\mathrm{k}}-\left(\frac{15}{3}\right)(\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}) \\
& =4 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+7 \hat{\mathrm{k}}-5(\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}) \\
& =-\hat{i}-8 \hat{j}+2 \hat{k}
\end{aligned}\)